FR06 · L3 · Chapter 7 of 9

Connect an initial shape to a final angular pattern

A shorter route through an almond-shaped pressure field can give a stronger outward push. The final particle distribution can therefore favour the initial short axis. To turn that picture into a calculation, we need separate numbers for the spatial shape and for the momentum-space angular pattern.

Before you beginYou need weighted averages, sine and cosine. The local pressure discussion in chapter 1 explains what accelerates matter; the transverse response below is a new reduced model, separate from the longitudinal ODE.

Sandbox Physics · 2026-10-27

Measure the shape about its own centre

Let $\rho(x,y)$ be the transverse deposited density. Recenter it so its density-weighted mean position vanishes. Define $\langle F\rangle_\rho=\int F\rho\,dxdy/\int\rho\,dxdy$ . A complex exponential is a rotating unit arrow: $e^{i\theta}=\cos\theta+i\sin\theta$ , with $i^2=-1$ . Its real and imaginary parts are the two Cartesian components. In polar coordinates the eccentricity vector for harmonic $n=2,3$ is:

\varepsilon_n e^{in\Psi_n}=-\frac{\langle r^n e^{in\varphi}\rangle_\rho}{\langle r^n\rangle_\rho}

The magnitude measures the corresponding deformation. The angle sets its orientation; the minus sign makes the elliptic direction point along the short axis for a smooth elongated profile.

For an axis-aligned Gaussian with widths $\sigma_x,\sigma_y$ and $\sigma_y>\sigma_x$ , this gives $\varepsilon_2=(\sigma_y^2-\sigma_x^2)/(\sigma_y^2+\sigma_x^2)$ . A third harmonic measures a triangular distortion and has its own orientation in a real event.

A second harmonic describes momenta, not the photograph

Write the particle’s momentum angle as $\phi$ . A harmonic modulation of its probability density is $p(\phi)=[1+2v_n\cos n(\phi-\Psi_n)]/(2\pi)$ . Integrating this density against the cosine gives $v_n$ ; its magnitude is an angular moment, not the spatial eccentricity.

v_n=\int_0^{2\pi}p(\phi)\cos n(\phi-\Psi_n)\,d\phi

Orthogonality removes the uniform term and the other Fourier harmonics. With several harmonics the total probability density must stay nonnegative.

The spatial short axis can become the preferred momentum direction. The two shapes represent different quantities, connected by an assumed response law.

For modest second- and third-harmonic deformations, a linear response is useful. Gardim and collaborators tested estimators of this kind against event-by-event hydrodynamics. Higher harmonics can need products of lower eccentricities; extending one linear coefficient to every harmonic would lose that physics.

Declare a response law before varying viscosity

The Lab defines $v_n=\kappa_n\varepsilon_n$ and the toy damping law below. Here $\kappa_0$ is an undamped response scale, $C_\eta=\eta/s_{\rm eq}$ is the viscosity parameter, and $RT$ is a dimensionless size–temperature product. The numerical factor $a_d=1.8$ belongs to this teaching model.

\kappa_n=\kappa_0\exp\!\left[-\frac{n^2a_d C_\eta}{RT}\right]

The stronger damping at higher harmonic number illustrates sensitivity to finer angular structure. This exponential is a specified surrogate, not a fitted event-level relativistic-fluid solution.

Hold the initial shape fixed and increase viscosity: the emission modulation weakens. Hold the observed modulation fixed and instead assume a larger initial deformation: stronger damping can compensate. The next chapters turn that degeneracy into a statistical inference problem.

Worked example

Compute a visible elliptic modulation

For $\sigma_x=2$ and $\sigma_y=3$ , the spatial eccentricity is $\varepsilon_2=5/13\simeq0.38462$ . Choose $\kappa_0=0.34$ , $C_\eta=0.16$ and $RT=3.5$ .

v_2=0.34\frac5{13}\exp\!\left[-\frac{4(1.8)(0.16)}{3.5}\right]\simeq0.0941

The density along the preferred axis is proportional to 1.188, and across it to 0.812. Their ratio is about 1.46. This angular excess is smaller than the initial spatial deformation.

Predict, then try the model

Does a larger viscosity increase or decrease the preferred-direction excess at fixed shape?

Compare An elliptic cloud with Almost round. Restore the elliptic cloud and increase viscosity with geometry fixed. Look at the emission directions and the angular distribution; the third harmonic is introduced separately in the frozen inference sample.

Shape to collective flow ↗

Compare your explanation

The second-harmonic excess weakens under viscous damping. The third harmonic obeys the same declared response law in the separate inference benchmark, with a larger squared harmonic factor.

Exercises with explanations

1. A spatial density is rotated without changing its shape. What should happen to the magnitudes of eccentricity and flow?

Read the reasoning

Their magnitudes stay the same. Their orientation angles rotate with the density. A model that changes the magnitudes under a rigid rotation would violate rotational symmetry.

2. Could an exactly circular smooth profile have triangular flow in this deterministic linear model?

Read the reasoning

It has zero triangular eccentricity and therefore zero triangular response. Event-by-event density fluctuations can supply a nonzero triangular input, which is extra initial-state information.

Sources and reading scope

Gardim et al., Mapping the hydrodynamic response to the initial geometry in heavy-ion collisions, arXiv:1111.6538v2
Version 2 (5 March 2012), §§II–IV: eccentricity estimators, linear response and nonlinear higher harmonics. The teaching exponential damping law is separately defined, not fitted from its event simulations.
Bilandzic, Snellings & Voloshin, Flow analysis with cumulants: direct calculations, arXiv:1010.0233v2
Version 2 (17 October 2011), Eqs. (3)–(4), (11)–(18): nonflow, Q vectors and exact distinct-particle correlators for uniform acceptance.